A metallic bucket is in the shape of frustum of a cone If the diameters of two circulear ends of - Maths - Surface Areas and Volumes

Question

A metallic bucket is in the shape of frustum of a cone If the
diameters of two circulear ends of the bucket are 45 and 25 cm
respectively and total height is 24cm find the area of metallic
sheet used to make the bucket Also find the volume

Manmeet Singh · Accepted Answer

Height of the bucket, h = 24 cm (Given)

Let the radii of the upper and lower ends of the bucket be r₁ and r₂respectively.

Given, r₁ = $\frac{45}{2} = 22.5 c m$ cm and r₂ = $\frac{25}{2} = 12.5 c m$ cm.

$Slant height of bucket, l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}} = \sqrt{{(24 cm)}^{2} + (22.5 cm - 12.5 cm)^{2}} = \sqrt{{(24 cm)}^{2} + (10 cm)^{2}} = \sqrt{576 {cm}^{2} + 100 {cm}^{2}} = \sqrt{676 {cm}^{2}} = 26 cm$

Surface are of bucket

= Curved surface area of the bucket + Area of base of the bucket

= π(r₁ + r₂)l + πr₂²

= π [ (r₁ + r₂)l + r₂² ]

= $\frac{22}{7}$ [ (22.5 cm + 12.5 cm) 26 cm + (12.5 cm)² ]

= $\frac{22}{7}$ (35 × 26 cm² + 156.25 cm²)

= $\frac{22}{7}$ ( 910 cm² + 156.25 cm² )

= $\frac{22}{7}$ × 1066.25 cm²

= 3351. 07 cm²

So, the area of metallic sheet used to make the bucket = 3351. 07 cm²

Volume =

$= \frac{1}{3} \times \frac{22}{7} \times 24 ({22.5}^{2} + {12.5}^{2} + 22.5 \times 12.5) = \frac{528}{21} (506.25 + 156.25 + 281.25) = \frac{528}{21} (943.75) = 23728.57 c m^{3}$

$= \frac{1}{3} \times \frac{22}{7} \times 24 ({22 5}^{2} + {12 5}^{2} + 22 5 \times 12 5) = \frac{528}{21} (506 25 + 156 25 + 281 25) = \frac{528}{21} (943 75) = 23728 57 c m^{3}$

A metallic bucket is in the shape of frustum of a cone.If the diameters of two circulear ends of the bucket are 45 and 25 cm respectively and total height is 24cm find the area of metallic sheet used to make the bucket.Also find the volume.